A study was performed to evaluate whether zidovudine (AZT) administered to HIV-i
ID: 3269653 • Letter: A
Question
A study was performed to evaluate whether zidovudine (AZT) administered to HIV-infected pregnant women and their newborns could reduce the rate of transmission from mother to infant. The final study population consisted of 363 live births. The mothers of 180 were randomly assigned to the AZT arm of the study and 13 of the infants developed HIV infections. The mothers of 183 were randomly assigned to receive a placebo and 40 of those infants developed HIV.
What type of study is this? : Randomized controlled trial
Construct a 2 X 2 table for these data.
HIV
No HIV
Total
AZT
13
167
180
Placebo
40
143
183
Total
53
310
363
c. (4 pts) Calculate the appropriate measure of association between the AZT administration to pregnant women and transmission of HIV to their babies and state what this means.
A study was performed to determine the risk of coronary heart disease (CHD) in individuals with hypertension. 1,387 men between the ages of 40 to 59 with no previous CHD, were categorized as normal (no hypertension), borderline hypertension or definite hypertension. These men were followed for six years and the incidence of new CHD was recorded. There were 556 men with normal blood pressure and of those, 23 developed CHD. There were 532 men with borderline hypertension and of those, 38 developed CHD. There were 299 with definite hypertension and of those, 37 developed CHD.
a. What type of study is this?
b. Construct a 2 X 3 table for these data
c. Calculate the appropriate measure of association between borderline hypertension and CHD and between definite hypertension and CHD. Use the risk seen in the normal (no hypertension) group as your reference and compare the remaining two groups to the normal group.
d. Interpret one of the numbers you calculated in part c.
HIV
No HIV
Total
AZT
13
167
180
Placebo
40
143
183
Total
53
310
363
Explanation / Answer
Solution
AZT - HIV-infection
Part (a)
Type of study: Randomized Controlled Trial
Part (b)
2 X 2 table for the given data.
HIV
No HIV
Total
AZT
13
167
180
Placebo
40
143
183
Total
53
310
363
Part (c)
Test of Association
Back-up Theory
Suppose we have a contingency table consisting of r rows and c columns, i.e., (r x c) cells, each row representing a particular level of one attribute and each column representing a particular level of another attribute. Let oij be the observed frequency in the cell at ith row-jth column and eij be the corresponding expected frequency.
H0: The two attributes are independent. vs H1: H0 is false.
Test Statistic
2 = (i = 1 to r, j = 1 to c){(oij - eij)2/eij}. [This can also be simplified as 2 = (i = 1 to r, j = 1 to c)(o2ij/eij) – n, where n = total frequency = (i = 1 to r, j = 1 to c)(oij) which is also equal to (i = 1 to r, j = 1 to c)(eij).]
Under H0, 2 is distributed as Chi-square distribution with degrees of freedom = (r - 1)(c - 1).
Under H0, eij = (oi. . o.j)/n where oi. = (j = 1 to c)(oij) = sum of all observed frequencies in the ith row and o.j = (i = 1 to r)(oij) = sum of all observed frequencies in the jth column.
Decision Criterion
Reject H0 if 2cal > 2(r - 1)(c - 1), . i.e. reject H0 if 2cal > 2crit (i.e., calculated value of 2 is greater than the upper % point of Chi-square distribution with degrees of freedom = (r - 1)(c - 1), which can be read off from standard table).
Now, to work out the solution,
In the given problem, the two attributes are: (AZT) administer and HIV-infection
r = 2, c = 2. Hence, degrees of freedom = 1 x 1 = 1
From the calculations shown below, 2cal = 15.60.
Using Excel Function, 2crit = 21,0.05 = 3.84
Since 2cal > 2crit, H0 is rejected at 5% level of significance implying that the two attributes in the given problem are not independent.
Conclusion
There is enough evidence to suggest that HIV infection is associated with AZT administration.
Excel Calculations
OIJ (given)
Oi1
Oi2
Oi.
O1j
13
167
180
O2j
40
143
183
O.j
53
310
363
Eij
Ei1
Ei2
Total
E1j
26.28099
153.719
180
E2j
26.71901
156.281
183
Total
53
310
363
CHECK: respective row totals, column totals and grand total of both Oij and Eij must be equal.
(Oij - Eij)^2/Eij
1
2
1
6.711495
1.147449
7.86
2
6.60147
1.128638
7.73
13.31297
2.276088
15.6
Hypertension and Risk of CHD
Part (a)
Type of study: Randomized Controlled Trial
Part (b)
2 X 3 table for the given data.
Hypertension
Normal Blood Pressure
Borderline
Hypertension
Definite
Hypertension
Total
CHD
23
38
37
98
No CHD
533
494
262
1289
Total
556
532
299
1387
Following the very same steps asin the previous case,
In the given problem, the two attributes are: Hypertension and CHD
r = 2, c = 3. Hence, degrees of freedom = 1 x 2 = 2
From the calculations shown below, 2cal = 20.103.
Using Excel Function, 2crit = 22,0.05 = 5.991
Since 2cal > 2crit, H0 is rejected at 5% level of significance implying that the two attributes in the given problem are not independent.
Conclusion
There is enough evidence to suggest that CHD is associated with hypertension.
Excel Calculations
Oij
DF =
2
1
2
3
Oi.
1
23
38
37
98
2
533
494
262
1289
O.j
556
532
299
1387
Eij
1
2
3
Total
1
39.28479
37.58904
21.1
98
2
516.7152
494.411
278
1289
Total
556
532
299
1387
Chi-square
1
2
3
Total
1
6.75056
0.004493
11.9
18.68236
2
0.513231
0.000342
0.91
1.420381
Total
7.263791
0.004835
12.8
0
20.10275
HIV
No HIV
Total
AZT
13
167
180
Placebo
40
143
183
Total
53
310
363
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.